Question

A bear spies some honey and takes off from rest at a rate of 2.00 m/s^2. If the honey is 9.00 m away, how fast will his snout be going at the moment of ecstasy?

Answer 1

Answer:

$\rm 3\sqrt{2}\; m \cdot s^{-1}$, which is approximately 4.24 m/s.

Explanation:

Apply the equation for constant acceleration:

$\displaystyle x = \frac{v^{2} - u^{2}}{2a}$,

where

• $x$ is the displacement,
• $v$ is the final velocity,
• $u$ is the initial velocity, and
• $a$ is the acceleration.

For this question,

• $x = \rm 9.00\; m$,
• $v$ the final velocity needs to be found,
• $u = 0$ for the bear started from rest, and
• $a = \rm 2.00\; m\cdot s^{-2}$.

Rearrange the equation:

$v^{2} = 2a\cdot x + u^{2}$,

$\begin{aligned}v &= \sqrt{2a\cdot x + u^{2}}\\ &= \sqrt{2\times 2.00\times 9.00}\\ &= \sqrt{18}\\&=3\sqrt{2}\\&\approx 4.24\;\rm m \cdot s^{-1} \end{aligned}$.